For sustained interference fringes in a double-slit experiment,the essential condition$(s)$ is/are:
$1$. Sources must be coherent.
$2$. The intensities of the two sources must be equal.
Here,the correct option$(s)$ is/are:

Two sources produce an interference pattern which is observed on a screen,at a distance $D$ from the sources. The fringe width is $2w$. If the distance $D$ is now doubled,the fringe width will be:

$A$ light source,which emits two wavelengths $\lambda_1=400 \ nm$ and $\lambda_2=600 \ nm$,is used in a Young's double slit experiment. If recorded fringe widths for $\lambda_1$ and $\lambda_2$ are $\beta_1$ and $\beta_2$ and the number of fringes for them within a distance $y$ on one side of the central maximum are $m_1$ and $m_2$,respectively,then
$(A)$ $\beta_2 > \beta_1$
$(B)$ $m_1 > m_2$
$(C)$ From the central maximum,$3^{\text{rd}}$ maximum of $\lambda_2$ overlaps with $5^{\text{th}}$ minimum of $\lambda_1$
$(D)$ The angular separation of fringes for $\lambda_1$ is greater than $\lambda_2$

In Young's double-slit experiment,the intensity at a point is $(1/4)$ of the maximum intensity. The angular position of this point is:

In Young's double slit experiment,if the source of light changes from orange to blue,then

$A$ double slit experiment is immersed in water of refractive index $1.33$. The slit separation is $1 \,mm$, and the distance between the slit and the screen is $1.33 \,m$. The slits are illuminated by light of wavelength $6300 \,Å$. The fringe width is: